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Find the limit \lim _{x \rightarrow a} f(x) where f(x)=\frac{\log [x]}{x} , where [ ]  indicates the Floor Function.. The given limits are 

\lim _{x \rightarrow \infty} \frac{\log (x-1)}{x}=\lim _{x \rightarrow \infty} \frac{1}{(x-1)}, \lim _{x \rightarrow \infty} \frac{\log x}{x}=\lim _{x \rightarrow \infty} \frac{1}{x}

Option: 1

0 


Option: 2

\infty


Option: 3

1


Option: 4

\pi


Answers (1)

best_answer

Note the following important points:

The Sandwich theorem also known as squeeze play theorem applicable for any three real valued functions f(x),g(x) and h(x)  that shares the same domain such that f(x)\leq g(x)\leq h(x) for \forall x 

  • in the domain of definition states the following: 

  • For some real value of a, if \lim _{x \rightarrow a} f(x)=l=\lim _{x \rightarrow a} h(x), then \lim _{x \rightarrow a} f(x)=l
  • The Floor Function indicates the greatest integer function [GIF] denoted mathematically as [P] for only real values. This function [P] rounds downs the real number having any fractional or decimal part (if any) to the nearest integral value less than the indicated number.

From the definition of GIF, the following is evident.

\begin{aligned} & x-1<[x] \leq x \\ & \log (x-1)<\log [x] \leq \log x \end{aligned} --------(i)

The provided limits are

\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\log (x-1)}{x}=\lim _{x \rightarrow \infty} \frac{1}{(x-1)} \\ & \lim _{x \rightarrow \infty} \frac{\log x}{x}=\lim _{x \rightarrow \infty} \frac{1}{x} \quad \ldots(\text { iii) } \end{aligned}

Now, multiply the inequality (i) by the reciprocal of

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